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Transcript
```Solutions to Homework Set 6
1) A group is simple if it has no nontrivial proper normal subgroups. Let G
be a simple group of order 168. How many elements of order 7 are there in
G?
Solution: Observe that 168 = 23 · 3 · 7. Every element of order 7 generates
a cyclic group of order 7 so let us count the number of such subgroups: By
the Sylow theorems, the number of subgroups of order 7 is ≡ 1 (mod 7) and
divides 168. The only options are 1 and 8. There cannot be only one Sylow
7-subgroup, since it would be normal, so there are 8 of them. Every two of
these subgroups intersects in a subgroup whose order divides 7, so it must
be the identity. Every other element in these subgroups must have order 7,
so the number of elements of order 7 is 6 · 8 = 48.
2) Let G be a group of order 231. Prove that the (unique) Sylow 11-subgroup
of G is in the center.
Solution: There are several ways to proceed. Here is a sketch of one proof:
By arguments similar to the previous problem, there is a unique Sylow 7subgroup L and a unique Sylow 11-subgroup M , and of course these are both
normal. Now if we can produce a normal subgroup N of order 3 · 7 = 21
then the set M N is actually a group (why?) of order 231 so it must be
G. Moreover, M must commute with N , since if m ∈ M, n ∈ n, then by
normality of M, N , we have that mnm−1 n−1 ∈ M ∩ N = {1} so mn = nm.
It is now easy to show that in fact M must commute with G (since every
g ∈ G can be written as m0 n0 ). So it suffices to find a normal subgroup of
order 21. To this end, let P be a Sylow 3-subgoup and consider P L. It is
easy to show that this is a subgroup of order 21. To show that its normal,
argue that its normalizer cannot have order 21 and then it must be all of G.
3) Prove that for every even integer p ≥ 2, there exists a constant c(p) such
that any one-distance set with respect to the Lp -norm in Rn has at most nc(p)
points.
Solution: Proceed as in class, by defining fi (x) = ||x−vi ||pp −δ p , where δ is the
distance, and vi is the incidence vector for set Ai . Then these polynomials
are linearly independent and they lie in a space spanned by 1 and xji for
1 ≤ i ≤ n and 1 ≤ j ≤ p. The numbe rof these is 1 + pn.
1
4) Let λ be a nonzero integer, J be the m × m all ones matrix, and D be
the m × m diagonal matrix with positive integer diagonal entries γ1 , . . . , γm .
Compute the determinant of A = λJ + D.
Solution: We use a trick called bordering: Extend A to an (m + 1) by (m + 1)
matrix A0 by adding a first rwo of all ones and a first column of all ones except
the top left entry. Clearly det(A) = det(A0 ). Subtract λ times the first row
from each row. Now use the diagonal entries to kill the first column and
create an upper triangular matrix. The determinant is now the product of
the diagonal. The result is
γ1 · · · γm
1
1
1+λ
+ ··· +
γ1
γm
2
!!
```
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